What is the slope of the tangent line of # (x/y-2)(xy-3)-e^y= C #, where C is an arbitrary constant, at #(-2,1)#?

1 Answer

The slope of the tangent line #m=(-9)/(e+2)=-1.90747#

Explanation:

We start from the given equation

#(x/y-2)(xy-3)-e^y=C#

Expand the left side of the equation

#(x/y)(xy)-(3x)/y-2xy+6-e^y=C#

#x^2-(3x)/y-2xy+6-e^y=C#

Differentiate both sides of the equation with respect to #x#

#d/dx(x^2-(3x)/y-2xy+6-e^y)=d/dx(C)#

#d/dx(x^2)-3*d/dx(x/y)-2*d/dx(xy)+d/dx(6)-d/dx(e^y)=0#

#2x-3((y*1-x*y')/y^2)-2*(xy'+y*1)+0-e^y*y'=0#

Substitute right away the value of #x=-2# and #y=1# from the point #(-2,1)#

#2x-3((y*1-x*y')/y^2)-2*(xy'+y*1)+0-e^y*y'=0#

#2(-2)-3((1*1-(-2)*y')/1^2)-2[(-2)y'+1*1]+0-e^1*y'=0#

Simplify

#2(-2)-3((1*1-(-2)*y')/1^2)-2[(-2)y'+1(1)]+0-(e^1)y'=0#

#-4-3((1+2y')/1)+4y'-2+0-(e^1)y'=0#

#-4-3-6y'+4y'-2-ey'=0#

Solve for #y'#

#-6y'+4y'-ey'=4+3+2#

#(-6+4-e)y'=9#

#y'=9/(-6+4-e)#

#y'=9/(-2-e)#

#y'=(-9)/(2+e)#

#y'=-1.90747#

God bless....I hope the explanation is useful.